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Consider a random variable $V$ with distribution function $F$ and density function $f$ with support $[\underline{v},\overline{v}]$, where $0\leq\underline{v}<\overline{v}$. The mean is $\mu$. Here $f$ is assumed to be differentiable on its support and log-concave (i.e. $\ln \circ f$ is a concave function).

I would like to prove (or disprove) that

$$\tag{1}\phi(\mu)=\mu F(\mu)^2-\int_{\mu}^{\overline{v}}F(v)[1-F(v)]dv\geq 0$$

for all such probability distributions.

I have not been able to find any examples of distributions for which inequality $(1)$ is violated (I have tried the beta and Kumaraswamy distributions on $[0,1]$ with many different combinations of parameters).

An example of an $f$ that is not log concave for which $\phi(\mu)<0$ would also be helpful (as I am not entirely sure whether the log-concavity is relevant).


Inequality $(1)$ can also be written as (using integration by parts):

$$\begin{align}&E[V\mid V\geq \mu]\cdot P(V\geq \mu)+\mu\cdot P[V\leq \mu]-E[\hat{V}\mid \hat{V}\geq \mu]\cdot P(\hat{V}\geq \mu)\\ &=\int_{\mu}^{\overline{v}}vf(v)dv+\mu F(\mu)-\int_{\mu}^{\overline{v}}v\hat{f}(v)dv\\\tag{2} &=\int_{\mu}^{\overline{v}}vf(v)dv+\mu F(\mu)-\int_{\mu}^{\overline{v}}v[2f(v)F(v)]dv\\ &=\int_{\mu}^{\overline{v}}vf(v)[1-2F(v)]dv+\mu F(\mu)\geq 0\end{align} $$

where $\hat{V}$ denotes the random variable that is the maximum of two independent copies of $V$, $\hat{F}$ given by $\hat{F}(v)=[F(v)]^2$ is its distribution function and $\hat{f}$ given by $\hat{f}(v)=2f(v)F(v)$ is its density function.


Some extra details:

Let $\phi(x)=g(x)-h(x)$ where $g$ and $h$ are given by

$$g(x)=\int_{x}^{\overline{v}}vf(v)dv+\mu F(x),\qquad \text{and} \qquad h(x)=\int_{x}^{\overline{v}}v[2f(v)F(v)]dv$$

Then inequality $(1)$ is $\phi(\mu)\geq 0$. Denoting the mean of $\hat{F}$ by $\hat{\mu}$, it is easy to see that:

$$\phi(\underline{v})=\mu-\hat{\mu}<0\quad \text{and}\quad \phi(\overline{v})=\mu>0$$

Also, $\phi$ is strictly increasing:

$$\phi'(x)=(\mu-x+2xF(x))f(x)\geq xF(x)f(x)>0,\quad \forall x\in(\underline{v},\overline{v})$$

where the first inequality follows from Markov's inequality.

It follows that there exists a unique $\bar{x}\in(\underline{v},\overline{v})$ such that $$\phi(x)\lesseqqgtr0\iff x \lesseqqgtr\bar{x}.$$

Additionally $$g'(x)\gtreqqless0\iff x \lesseqqgtr\mu$$ so that the maximum of $g$ is $g(\mu)$.

Note also that integrating by parts gives

$$g(x)=\bar{v}+(\mu-x)F(x)-\int_x^{\overline{v}}F(v)dv\qquad \text{and}\qquad h(x)=\bar{v}-x[F(x)]^2-\int_x^{\overline{v}}[F(v)]^2dv$$


Some facts related to log-concavity:

  • Log-concavity of $f$ implies log-concavity of $F$ and $1-F$
  • If $f$ is differentiable then log-concavity of $F$ is equivalent to $f^2-Ff'\geq 0$ and log-concavity of $1-F$ is equivalent to $(1-F)f'+f^2\geq 0$
  • Log-concavity of $F$ implies $F(\mu)\geq \frac{1}{e}$
  • Products of log-concave functions are log concave. For example $F[1-F]$ is log-concave. -Log-concave densities are (strongly) unimodal

A proof that $\phi(\mu)\geq 0$ if $\mu\geq m$ and $\mu\geq \overline{v}/2$

Let $m$ be the median of $F$. Then

$$\phi(m)=\frac{1}{4}(2\mu-m)-\int_m^{\overline{v}}F(v)[1-F(v)]dv\geq \frac{1}{4}(2\mu-\overline{v})$$

where the last inequality follows because the integrand is at most $1/4$. Since $\phi$ is increasing and $\mu\geq m$, the result follows.

(Note that this result allows us to conclude that $\phi(\mu)\geq 0$ for all symmetric distributions.)


A lower bound for $\phi(\mu)$

From Chebyshev's integral inequality:

$$\int_{\mu}^{\overline{v}}[F(v)]^2dv\geq \frac{1}{\overline{v}-\mu}\left[\int_{\mu}^{\overline{v}}F(v)dv\right]^2$$

Thus

$$-\int_{\mu}^{\overline{v}}F(v)[1-F(v)]dv\geq -\frac{1}{\overline{v}-\mu}\int_{\mu}^{\overline{v}}F(v)dv\left(\overline{v}-\mu-\int_{\mu}^{\overline{v}}F(v)dv\right)$$

Since $\int_{\mu}^{\overline{v}}F(v)dv\leq \overline{v}-\mu$ and

$$\int_{\mu}^{\overline{v}}F(v)dv=\overline{v}-\mu F(\mu)-\int_{\mu}^{\overline{v}}vf(v)dv$$

we get

$$-\int_{\mu}^{\overline{v}}F(v)[1-F(v)]dv\geq \mu[1-F(\mu)]-\int_{\mu}^{\overline{v}}vf(v)dv.$$

It follows that $$\begin{align*} \phi(\mu)&\geq\mu F(\mu)^2+\mu[1-F(\mu)]-\int_{\mu}^{\overline{v}}vf(v)dv \\ &=\mu-\int_{\mu}^{\overline{v}}vf(v)dv-\mu F(\mu)[1-F(\mu)]\\ \tag{3} &=\int_{\underline{v}}^{\mu}vf(v)dv-\mu F(\mu)[1-F(\mu)]\\ &=\mu F(\mu)^2-\int_{\underline{v}}^{\mu}F(v)dv \end{align*}$$

The lower bound above seems to be nonnegative for the beta distributions and Kumaraswamy distributions. It is exactly zero for the uniform distribution on $[0,1]$.


A proof that $\phi(\mu)\geq 0$ when $f$ is increasing on $[\underline{v},\mu]$ and $F(\mu)\geq 1/2$

Since $F(\mu)[1-F(\mu)]\leq 1/4$, the lower bound $(3)$ gives the following sufficient condition for $\phi(\mu)\geq 0$: $$\int_{\underline{v}}^{\mu}vf(v)dv\geq\frac{1}{4}\mu$$

If $f$ is increasing on $[\underline{v},\mu]$ and $F(\mu)\geq 1/2$ then,

$$\int_{\underline{v}}^{\mu}vf(v)dv\geq \frac{\mu+\underline{v}}{2}F(\mu)\geq \frac{1}{4}\mu$$

where the first inequality comes from by Chebyshev's integral inequality.

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    $\begingroup$ Remark: The stronger inequality $$\mu F(\mu)^2 \geq \int_{\underline v}^{\overline v} F(x)-F(x)^2$$ is not true. For example, for a uniform distribution between $a=\underline v$ and $b=\overline v$, the LHS is $\frac{a+b}8$, but the RHS is $\frac{b-a}6$. However, your inequality might still be true... $\endgroup$ Commented Oct 19, 2019 at 18:31
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    $\begingroup$ Some observations: I think we can assume WLOG that $a=0$ (since the RHS is invariant under "shifts" and the LHS just gets larger when shifted); I tested the the inequality for all of the distributions here that satisfy your conditions (for the distributions with $n$ parameters I used at least $10^n$ parameter tuples). None of them gave a counter-example to your inequality. Lastly, I thought about using Hölder but there might be some long calculations. $\endgroup$ Commented Oct 20, 2019 at 0:38
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    $\begingroup$ @smcc I didn't get anything interesting using Hölder as is on the RHS. About your lower bound for $\phi(\mu)$ and general densities: Setting (here I use Iverson brackets) $a=0,b=1$ with $$F(x)=\ln(1+(e-1)\cdot x) \cdot [0\le x\le 1] + [1<x],$$ we get $\mu = \frac{e-2}{e-1}$ and so $F(\mu)=\ln(e-1)$ and $$\int_0^{\mu} F(x)\,\mathrm dx=-1 + \frac1{e-1}+\ln(e - 1)$$ so that $$\mu F(\mu)^2-\int_a^\mu F(x)\,\mathrm dx\approx-0.0008<0$$ (larger negative values can be reached by choosing $F$ to be more concave). NOTE THAT MY PDF IS NOT LOG-CONCAVE ON $[0,1]$. $\endgroup$ Commented Oct 22, 2019 at 15:36
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    $\begingroup$ @MaximilianJanisch Thanks for the useful example. For your example, $F$ and $1-F$ are log-concave (but as you say $f$ is log-convex). Also, $\phi(\mu)>0$. However, as you have calculated, my lower bound on $\phi(\mu)$ is negative. If $\phi(\mu)>0$ is true for all log-concave $f$, then my lower bound could still be enough. However, I cannot see a way to show it is nonnegative more generally than the restrictive conditions I have already. Your example at least shows that I would need to use log-concavity of $f$ in some way (and not just log-concavity of $F$ and $1-F$). $\endgroup$
    – smcc
    Commented Oct 22, 2019 at 16:43
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    $\begingroup$ @smcc I asked this question also on MathOverflow $\endgroup$ Commented Oct 22, 2019 at 23:32

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